Find a formula for the quadratic function whose graph has its vertex at (3,6) and its y -intercept at y = − 4 .

Answer 1

parabola with a vertex at (3,6) is

y-h=a(x-k)^2 with (h,k)=(3,6)

so

y-6 = a(x-3)^2

y= a(x-3)^2 +6 = ax^2 -6ax +9a + 6

if the y-intercept = -4 then 9a+6=-4

9a =-4+6 = 2

a = 2/9

y= (2/9)(x-3)^2 + 6 is a quadratic function with vertex (3,6) and y intercept y=-4

but that's an upward opening parabola

You could also have a downward sloping parabola with the same vertex

or a right opening or left opening parabola with that same vertex

so that's 4 possible quadratics that satisfy the (3,6) vertex and -4 y intercept

You could also rotate the parabola to any degree and get an infinite number of quadratics that satisfy the same vertex and y intercept requirements.